What is the arc length of f(x) = sinx on x in [pi/12,(5pi)/12] ?

1 Answer
Mar 13, 2018

L=pi/3+sum_(n=1)^oo((1/2),(n))1/4^n{((2n),(n))pi/3+2sum_(k=0)^(n-1)((n),(k))(sin((n-k)pi/3)cos((n-k)pi/2))/(n-k)}

Explanation:

f(x)=sinx

f'(x)=cosx

Arc length is given by:

L=int_(pi/12)^((5pi)/12)sqrt(1+cos^2x)dx

For x in [pi/12,(5pi)/12], cos^2x<1. Take the series expansion of the square root:

L=int_(pi/12)^((5pi)/12)sum_(n=0)^oo((1/2),(n))cos^(2n)xdx

Isolate the n=0 case:

L=int_(pi/12)^((5pi)/12)dx+sum_(n=1)^oo((1/2),(n))int_(pi/12)^((5pi)/12)cos^(2n)xdx

Apply the Trigonometric power-reduction formula:

L=[x]_ (pi/12)^((5pi)/12)+sum_(n=1)^oo((1/2),(n))int_(pi/12)^((5pi)/12){1/4^n((2n),(n))+2/4^nsum_(k=0)^(n-1)((n),(k))cos(2(n-k)x)}dx

Simplify:

L=pi/3+sum_(n=1)^oo((1/2),(n))1/4^nint_(pi/12)^((5pi)/12){((2n),(n))+2sum_(k=0)^(n-1)((n),(k))cos(2(n-k)x)}dx

Integrate directly:

L=pi/3+sum_(n=1)^oo((1/2),(n))1/4^n[((2n),(n))x+sum_(k=0)^(n-1)((n),(k))sin(2(n-k)x)/(n-k)]_(pi/12)^((5pi)/12)

Simplify:

L=pi/3+sum_(n=1)^oo((1/2),(n))1/4^n{((2n),(n))pi/3+sum_(k=0)^(n-1)((n),(k))(sin((n-k)(5pi)/6)-sin((n-k)pi/6))/(n-k)}

Apply the Trigonometric sum-to-product identity sina-sinb=2sin((a-b)/2)cos((a+b)/2):

L=pi/3+sum_(n=1)^oo((1/2),(n))1/4^n{((2n),(n))pi/3+2sum_(k=0)^(n-1)((n),(k))(sin((n-k)pi/3)cos((n-k)pi/2))/(n-k)}

Isolate the n=1 case:

L=pi/3+pi/12+sum_(n=2)^oo((1/2),(n))1/4^n{((2n),(n))pi/3+2sum_(k=0)^(n-1)((n),(k))(sin((n-k)pi/3)cos((n-k)pi/2))/(n-k)}